Question

Calculate the molarity and molality of nitric acid in a sample which has a density , 1.41 g/ml and the mass percent of nitric acid in it being 69%

Answer 1

Mass percent of nitric acid in the sample = 69 % (Given)

Thus, 100 g of nitric acid contains 69g of nitric acid by mass.

Molar mass of nitric acid (HNO3)

= {1 + 14 + 3(16)} g mol–1

= 1 + 14 + 48

= 63g mol–1

Therefore, number of moles in 69 g of HNO3 = 69g/63g mol–1

= 1.095 mol

Volume of 100g of nitric acid solution = Mass of solution/Density of solution

= 100g/1.41gmL–1

= 70.92mL = 70.92×10-³L

Concentration of nitric acid = 1.095 mol/ 70.92×10-³L

                                              = 15.44mol/L